Math 20-1 Chapter 9 Linear and Quadratic Inequalities

9.3 Quadratic Inequalities in Two Variables

Teacher Notes

To graph a quadratic inequality in 2 variables:

1. Graph the boundary parabola: solid or dashed

2. Shade the appropriate region: inside or outside

Remember: Any point in the shaded region is a solution

2y ax bx c 2y ax bx c 2y ax bx c 2y ax bx c

A quadratic inequality in two variables is an inequality that can be written in one of the forms below, where a, b, and c are real numbers and a ≠ 0.

9.3 Quadratic Inequalities in Two Variables

9.3.1

Quadratic Inequalities in Two Variables2Graph the solution to ( 2) 1y x

Graph the related equation

y = (x – 2)2 + 1

Why do you use a solid line for the curve?

Choose a test point.

y > (x – 2)2 + 10 > (0 – 2)2 + 10 > 4+ 10 > 5 False

(0, 0)

(0, 0)

The chosen test point is outside of the parabola. Since this test point does not satisfy the inequality, shade inside of the parabola.

9.3.2

2 4y x x

Shade below (inside) the parabola because the solution consists of y-values less than those on the parabola for corresponding x-values.

Quadratic Inequalities in Two Variables2Graph the solution to 4y x x

Graph the related equation

2 4y x x 2 4y x x

Check using a test point.(–2, 0).

(–2, 0)

2

20 2

4

( ) 24( )

0 4 4

0 0

y x x

9.3.3

Graphing a Quadratic InequalityChoose the correct shaded region to complete the

graph of the inequality.

9.3.4

Graph each inequality.

1. 2. 2 2 8y x x 22 3 1x x y

Your Turn

9.3.5

Quadratic Inequality in Two Variables

22y x

2 4 6y x x

2 6 5y x x

22y x 2y x

2 6 10y x x

Match each inequality to its graph.

y x x 2 6 5

y x 22 , y x 22 , y x 2

y x x 2 4 6, y x x 2 6 10,

9.3.6

Quadratic Inequality in Two Variables

Light rays from a flashlight bulb bounce off a parabolic reflector inside a flashlight. The reflected rays are parallel to the axis of the flashlight. A cross section of a flashlight’s parabolic reflector is shown in the graph.Determine the inequality that represents the reflected light, if the vertex of the parabola is at the point (0, 1).

Since the vertex is at (0, 1) use y = a(x – p)2 + q.

Substitute p = 0 and q = 1.

y = a(x – 0)2 + 1y = a(x)2 + 1.

Use the point (15, 10) to solve for a.

y = a(x)2 + 110 = a(15)2 + 11

25a

Since the shaded region is above the parabola with a solid line, the inequality is

211

25y x

9.3.7

Your TurnDetermine the equation of the given

inequality.

Quadratic Inequality in Two Variables

Since the vertex is at (1, 5) use y = a(x – p)2 + q.

Substitute p = 1 and q = 5.

y = a(x – 1)2 + 5y = a(x – 1)2 + 5.

Use the point (2, 3) to solve for a.

y = a(x – 1)2 + 5 3 = a(2 – 1)2 + 5–2 = a

Since the shaded region is below the parabola with a broken line, the inequality is

y < –2(x – 1)2 + 5

9.3.8

For the photo album you are making, each page needs to be able to hold 6 square pictures. If the length of one side of each picture is x inches, then A ≥ 6x2 is the area of one album page. a) Graph this function.b) If you have an album page that has an area of 70 square inches, will it be

able to accommodate 6 pictures with 3-inch sides?

Quadratic Inequality in Two Variables

A = 6x2

Length of side

Are

a

a) Graph the related equation.

A = 6x2

Shade above the line

A ≥ 6x2

b) Plot the point (3, 70)

(3, 70)

Since the point (3, 70) lies within the solution region, the album page will accommodate the 6 pictures.

9.3.9

9.3.10

Assignment

Suggested QuestionsPage 496:1a, 3, 6, 7, 9, 10, 11, 12, 16